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Metamath Proof Explorer

Theorem coeq1d

Description: Equality deduction for composition of two classes. (Contributed by NM, 16-Nov-2000)

		Ref	Expression
	Hypothesis	coeq1d.1	$⊢ φ \to A = B$
	Assertion	coeq1d	$⊢ φ \to A \circ C = B \circ C$

Step	Hyp	Ref	Expression
1		coeq1d.1	$⊢ φ \to A = B$
2		coeq1	$⊢ A = B \to A \circ C = B \circ C$
3	1 2	syl	$⊢ φ \to A \circ C = B \circ C$